Kshm2686
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Could anyone help me with these questions for 5. I) I think the answer would be domain: x is any real number and range: y is any real number. if it’s not right can anyone explain it please🙏🏼. For part ii) do you just sub the value into your original f(x)? and for iii) I know how to do this one

for question 6 I) I got as far as 4sec^2xcosec^2x-2sec^2x-2cosec^2x+1 and I don’t know how to continue and for part ii) I have no idea.

Any help would be appreciated 🙏🏼🙏🏼
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Kshm2686
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RDKGames
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(Original post by mantekes)
Could anyone help me with these questions for 5. I) I think the answer would be domain: x is any real number and range: y is any real number. if it’s not right can anyone explain it please🙏🏼.
Neither domain nor range are correct.

I suggest you first study the domain and range of \cos^{-1}(x) before moving onto this questions function.

for question 6 I) I got as far as 4sec^2xcosec^2x-2sec^2x-2cosec^2x+1 and I don’t know how to continue
Im not sure how you got this. Post your working out??

and for part ii) I have no idea.

Any help would be appreciated 🙏🏼🙏🏼
Combine these two fractions into a single one. Im sure you know how to add fractions with different denominators?
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ghostwalker
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(Original post by mantekes)
for question 6 I) I got as far as 4sec^2xcosec^2x-2sec^2x-2cosec^2x+1 and I don’t know how to continue and for part ii) I have no idea.
Q6.

Not sure how you arrived at that, but it seems OK, for the lefthand side.

Comparing what you have there, with what you're trying to get to on the right, you need to show that 2\sec^2x\operatorname{cosec}^2x-2\sec^2x-2\operatorname{cosec}^2x=0

One way would be to rewrite them in terms of sines and cosines and put them over a common denominator. It simplifies very quickly.

For the second part, I'd rewrite the tan and cot in terms of sine and cosine to start, then simplify the denominators, etc.

Edit: Too slow, but I'm not going to waste those \operatornames.
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Physics Enemy
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(Original post by mantekes)
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5) i) Domain of 2cos^-1 is range of 2cos, which is ... so domain of f(x) is? Range of 2cos^-1 is domain of 2cos, ensuring 1-1 map, thus ... so domain of f(x) is?

ii) Let y = 2cos^-1(-0.5) => cos(y/2) = -0.5
=> y/2 = ... => y = ... so f(-0.5) is?

6 i) LHS = (1 + 2tan^2)(1 + 2cot^2)
= 1 + 2(tan^2 + cot^2 + 2)
= 1 + 2[sin^4 + cos^4 + 2(sincos)^2](sec.cosec)^2
= ...

ii) Note: 1 - cot = -cot(1 - tan)
LHS = (cos - tansin)/(1 - tan)
= (cos - tan^2cos)/(1 - tan)
= cos(1 - tan^2)/(1 - tan)
= ...
Last edited by Physics Enemy; 1 year ago
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Kshm2686
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Thanks everyone!
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